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QUESTION

Consider the IVP $y' = te^{3t} - 2y$ for $0 \le t \le 1$ with $y(0) = 0$

and actual solution

$$y(t) = \frac{1}{5}te^{3t} - \frac{1}{25}e^{3t} + \frac{1}{25}e^{-2t}$$

(a) Use the Adams-Bashforth Three step method to approximate the solution, h=0.2, to the IVP. Use exact starting values.

(b) Use the Adams-Bashforth Three step method to approximate the solution, with $h=0.2$, to the IVP.

ATTEMPT

The Adams-Bashforth method looks as follows:

$$y_{i+1} = y_i + \frac{h}{12} (23f(t_i, y_i) - 16f(t_{i-1}, y_{i-1}) + 5f(t_{i-2},y_{i-2}))$$

I know we're supposed to use the Runge-Kutta method to find out what $$(23f(t_i, y_i) - 16f(t_{i-1}, y_{i-1}) + 5f(t_{i-2},y_{i-2}))$$ is. I don't know how. I only know the Runge-Kutta steps to find out $y_{i+r}$, $r\in N$

I'm not sure how to use it to find out $y_{i-r}$, where $r \in N$

PLEASE HELP

(Not homeweork. Studying for a test)

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  • $\begingroup$ A sketch: you have the initial conditions, and you can call the point your $(t_0,y_0)$. You then use a Runge-Kutta method (the classical fourth order one is standard, but a third-order version would be more suitable) to produce $(t_1,y_1)$ from your initial value. From this new point, use RK again to get $(t_2,y_2)$. Now that you have three starting points, you can then use Adams-Bashforth, using those three pairs you now have. $\endgroup$ – J. M. is a poor mathematician Apr 14 '13 at 11:20
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    $\begingroup$ On the other hand, if you're expected to use exact starting values, since you seem to already have the exact answer anyway, evaluate your $y(t)$ at the points $t=0,0.2,0.4$, and then use those three points to load up AB. $\endgroup$ – J. M. is a poor mathematician Apr 14 '13 at 11:23
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We are given the IVP:

$\tag 1 y' = te^{3t} - 2y$ for $0 \le t \le 1$ with $y(0) = 0$

and actual solution

$\tag 2 y(t) = \frac{1}{5}te^{3t} - \frac{1}{25}e^{3t} + \frac{1}{25}e^{-2t}$

(a) Use the Adams-Bashforth Three step method to approximate the solution, h=0.2, to the IVP. Use exact starting values.

Exact (Using the exact solution and time steps given by h.)

  • $y(0.0) = 0$
  • $y(0.2) = 0.0268128$
  • $y(0.4) = 0.150778$
  • $y(0.6) = 0.49602$
  • $y(0.8) = 1.33086$
  • $y(1.0) = 3.2191$

Using the Adams-Bashforth three-step method, we have:

$w_0 = 0$

$w_1 = 0.0.0268128$

$w_2 = 0.150778$

$\displaystyle w_{i+1} = w_i + \frac{h}{12} (23f(t_i, w_i) - 16f(t_{i-1}, w_{i-1}) + 5f(t_{i-2},w_{i-2}))$

Can you take it from here for the exact method?

(b) Use the Adams-Bashforth Three step method to approximate the solution, with $h=0.2$, to the IVP.

For this case, we need to use the Runge-Kutta to find the starting values.

We would use the Third Order Runge - Kutta

Using the 3rd-order RK For the starting values, we would set:

  • $t = a = 0$, $w_0 = \alpha = 0$, and then for $i= 1, 2, \ldots, N$ do:

  • $\displaystyle k_1 = f(t, w)$

  • $\displaystyle k_2 = f(t + \frac{1}{2} h, w + \frac{1}{2}k_1 h)$
  • $\displaystyle k_3 = f(t + h, w -k_1 h + 2k_2 h)$
  • $\displaystyle w_i = w_{i-1} + \frac{1}{6}\left(k1 + 4 k_2 + k_3\right)h$
  • $\displaystyle t = a + i h$

After we have those starting values, we would proceed as before with the Adams-Bashforth (just have starting values that used Runge-Kutta as opposed to exact).

Does that all make sense?

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